How to prove this inequality in interval: (0,π2) 2cos⁡x+sec2x−3≥0

oliviayychengwh

oliviayychengwh

Answered

2022-01-02

How to prove this inequality in interval: (0,π2)  2cosx+sec2x30

Answer & Explanation

Thomas Nickerson

Thomas Nickerson

Expert

2022-01-03Added 32 answers

Restrict the domain to some meaningful interval such as x(π2,π2) so that cosx>0. In that case, using the AM-GM inequality you have:
LHS=cosx+cosx+sec2x33cosxcosxsec2x33=33=0=RHS
Papilys3q

Papilys3q

Expert

2022-01-04Added 34 answers

We have
2cosx+sec2x3=2(cosx1)+(1cos2x1)=(2cosx+1)(cosx1)2cos2x
Thefore the inequality does not hold.
Vasquez

Vasquez

Expert

2022-01-09Added 457 answers

By Fermat's theorem, we examine zeroes of derivative of the function f(x)=2cosx+sec2x3 to find the extrema, that is, to solve the following:
f(x)=2tanxsec2x2sinx=0
The solution is x={kπ|kZ}. When x=π, we get
2cos(x)+sec2(x)3=4
Thus 2cosx+sec2x30 doesn't hold.
Since the interval is added, we can prove the claim is true. For x=0, f(x)=0. And when x(0,π2), it is trivial that f(x)0, thus f(x) monotonously increases on (0,π2). Hence f(x)f(0)=0 for all x(0,π2) and we finished the proof.

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